every even integer greater than 70 is the sum of two abundant numbers in more than one way

Theorem Every even n>70 can be expressed as n=a+b, with both a and babundant numbers, in more than one way. Due to the commutative property of addition, swaps of a and b are not counted as separate ways.

Proof.

To prove this it is enough to find just two ways for each even n>70, though of course there are plenty more ways as the numbers get larger, purely for our convenience we’ll seek to choose the smallest values for b possible. Since every multiple of a perfect number is an abundant number, and 6 is a perfect number, it follows that every multiple of 6 is abundant, and it is small enough a modulus that reviewing all possible cases should not prove tiresome.

If n=6⁢m and m>5, the two desired pairs are a=6⁢(m-2), b=12, and a=6⁢(m-3), b=18. This leaves us the cases n=6⁢m+2 and n=6⁢m+4 to concern ourselves with.

If n≡2mod6 and m>10 then the pairs are are a=6⁢(m-3), b=20, and a=6⁢(m-9), b=56.

If n≡4mod6 and m>12 then the pairs are a=6⁢(m-6), b=40, and a=6⁢(m-11), b=70.

The lower bounds of m have been chosen to ensure the formulas give distinct pairs of abundant numbers and never the perfect number 6 itself, but its multiples. These values of m correspond to the values of n 36, 68, 82. To complete the proof we are left with the special case of n=76 to examine on its own. Ignoring the bounds for m, the formulas above give us 76 = 40 + 36, a valid pair, and 76 = 70 + 6, which is not a pair of abundant numbers. But there is one other pair, 56 + 20, of which neither a nor b is a multiple of 6.
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The special case of 76 shows that there are solutions that don’t use multiples of 6. These become more readily available as the numbers get larger.

Title

every even integer greater than 70 is the sum of two abundant numbers in more than one way